The equation sqrt{(x - 2)^2 + y^2} + sqrt{(x | vedclass

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(B) The given equation is $\sqrt{(x - 2)^2 + y^2} + \sqrt{(x + 2)^2 + y^2} = 4$. This represents the sum of distances from a point $P(x, y)$ to two fi

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Pair of straight lines

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(B) The given equation is $\sqrt{(x - 2)^2 + y^2} + \sqrt{(x + 2)^2 + y^2} = 4$. This represents the sum of distances from a point $P(x, y)$ to two fixed points $F_1(2, 0)$ and $F_2(-2, 0)$ being equal to $4$. The distance between the fixed points $F_1$ and $F_2$ is $d = \sqrt{(2 - (-2))^2 + (0 - 0)^2} = \sqrt{4^2} = 4$. Since the sum of the distances from $P$ to $F_1$ and $F_2$ is equal to the distance between $F_1$ and $F_2$ (i.e.,$PF_1 + PF_2 = F_1F_2$),the point $P$ must lie on the line segment connecting $F_1$ and $F_2$. This segment lies on the $x$-axis where $y = 0$ and $-2 \le x \le 2$. However,in the context of conic sections,this degenerate case represents a line segment,which can be considered as a pair of coincident straight lines $y = 0$.

The equation $\sqrt{(x - 2)^2 + y^2} + \sqrt{(x + 2)^2 + y^2} = 4$ represents a

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